The trace of a linear map $A : V \to V$ is always the same as the trace of its restriction to the unique largest vector subspace $W$ of $V$ such that $A\big\vert_W : W \to W$ is an isomorphism. By using this fact, any interpretation of the trace of linear *iso*morphisms can be extended to an interpretation of *all* linear endomorphisms (even if they are not isomorphisms).

Below is a proof that $\operatorname{Tr} A = \operatorname{Tr}\left(A\big\vert_W\right)$, where $W$ is the unique largest vector subspace of $V$ such that $A\big\vert_W : W \to W$ is an isomorphism and $A\big\vert_W$ has codomain $W$.

To begin, let
$$V^{(0)} = \operatorname{domain}(A) = V,\;\; V^{(i+1)} = A\left(V^{(i)}\right),\; \textrm{ and }d^i = \dim V^{(i)}$$
so that $V^{(1)} = \operatorname{Im} A = A\left(V^{(0)}\right)$, $V^{(i+1)} \subseteq V^{(i)}$, and $d^{i+1} \leq d^i$. Let $N \geq 0$ be the smallest integer s.t. $d^{N+1} = d^N$ and denote this common value by $d$. Let $W := V^{(N)}$.

We prove below that the restriction $A\big\vert_W : W \to W$ of $A$ onto $W := V^{(N)}$ is an isomorphism. Furthermore, $\operatorname{Tr}(A) = \operatorname{Tr}\left(A\big\vert_W\right)$ and it will be clear that $W$ is the unique largest vector subspace $S$ of $V$ on which $A$ restricts to an isomorphism $A\big\vert_S : S \to S$. All of this allows us to conclude that to geometrically interpret $\operatorname{Tr}(A)$, one may restrict their focus to geometrically interpreting the trace of the isomorphism $A\big\vert_W : W \to W$ rather than $A : V \to V$ itself.

This isn't entirely surprising since just as the trace of a matrix does not depend on the "elements off the diagonal", so too does the geometric interpretation of trace not depend on the "space off of $W$." This also gives some geometric intuition about how the trace of a matrix can simultaneously depend only on its diagonal elements while also equaling quantities that non-trivially depend on the whole matrix (such as the sum of its eigenvalues).
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**Proof:** We now prove the above claim. Inductively construct a basis $\left(e_1, \dots, e_{\dim V}\right)$ for $V$ such that for all $i \geq 0$, $\left(e_1, \dots, e_{d^i}\right)$ is a basis for $V^{(i)}$. Let $\left(\varepsilon^1,\dots, \varepsilon^{\dim V}\right)$ be the dual basis of $e_{\bullet}$ and note in particular that:
$$\textrm{(1) whenever }d^{i + 1} < l \leq d^i\textrm{ then }\varepsilon^l\textrm{ vanishes on }V^{(i + 1)}.$$

Since $(e_1, \dots, e_{d^1})$ is a basis for the range of $A$ we may, for any $v \in V^{(0)},$ write
$$A(v) = \varepsilon^1(A(v)) e_1 + \cdots + \varepsilon^{d^1}(A(v)) e_{d^1}$$
so that $A = (\varepsilon^l \circ A) \otimes e_l$ (the sum ranging over $l = 1, \dots, d^1$) and hence
$$\operatorname{Tr}(A) = (\varepsilon^l \circ A)(e_l) = \varepsilon^1(A(e_1)) + \cdots + \varepsilon^{d^1}\left(A\left( e_{d^1} \right)\right)$$
which shows that $\operatorname{Tr}(A)$ actually depends only on the range of $A$ (i.e. $V^{(1)}$).
Now since $e_1, \dots, e_{d^1}$ are (by definition) in $V^{(1)}$, all of $A\left(e_1\right), \dots, A\left(e_{d^1}\right)$ belong to $A\left(V^{(1)}\right) = V^{(2)}$ so that from $(1)$ it follows that
$$\operatorname{Tr}(A) = \varepsilon^1\left(A\left(e_1\right)\right) + \cdots + \varepsilon^{d^2}\left(A\left( e_{d^2} \right)\right)$$

Continuing this inductively $N \leq \dim V$ times shows that
$$\operatorname{Tr}(A) = \varepsilon^1\left(A\left(e_1\right)\right) + \cdots + \varepsilon^{d}\left(A\left(e_d\right)\right)$$
so that $\operatorname{Tr}(A)$ depends *only* on $W = V^{(N)}$.
Since by definition of $N$, the map $A\big\vert_W : W \to W$ is surjective, it is an isomorphism and furthermore, it should be clear that $W$ is the unique largest subspace of $V$ on which $A$ restricts to an isomorphism. $\blacksquare$

As described elsewhere, if you view $A : V \to V$ as a vector field on $V$ in the canonical way then the trace of $A$ is the same as its divergence so in the case where $A$ is an isomorphism there is a pleasing geometric interpretation readily available, which I'll assume that you're comfortable with. In my opinion, the equality $\operatorname{div}(A) = \operatorname{Tr}(A)$ is our best bet at finding a geometric interpretation of trace since it establishes a direct simple relationship between the trace and a readily interpretable quantity: $\operatorname{div}(A)$. An explanation of how this interpretation can be extended to linear maps that are not isomorphisms is now given.

Let $A : V \to V$ be an arbitrary linear map. Starting with the space $V = V^{(0)}$, imagine $A$ as transforming this space into $V^{(1)} := A\left(V^{(0)}\right)$. Then use $A$ transform $V^{(1)}$ into $V^{(2)} := A\left(V^{(1)}\right)$, and continuing transforming these spaces until eventually (after $N$ iterations) $A$ no longer transforms $V^{(N)}$ into a space with a strictly smaller dimension; that is: $\operatorname{dim} V^{(N)} = \operatorname{dim} A\left(V^{(N)}\right)$.
It is at this point that $A$ does nothing more than isomorphically transform the vector space $W := V^{(N)}$.
It is now possible apply your favorite interpretation of "trace of an isomorphism" to the isomorphism $A\big\vert_W : W \to W$, which then becomes an interpretation of the trace of the original linear map $A$ via $\operatorname{Tr}(A) = \operatorname{Tr}\left(A\big\vert_W\right) = \operatorname{div}\left(A\big\vert_W\right)$ represents.

Remark: This may not really answer your question since you stated that "The divergence application of trace is somewhat interesting, but again, not really what we are looking for." Nevertheless, whatever alternative non-divergence based interpretation of the trace of an isomorphism you choose, I hope that this will help you to extend it to the case where the map isn't an isomorphism.

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